Acceleration of a rod when one end is released

[indie452]

Homework Statement

two people are holding the ends of a rod length l and mass M, show that if one person let's go the initial acceleration of the free end is 3g/2

The Attempt at a Solution

i worked out the moment of inertia about centre mass (cm) and got = Ml²/12

because L=Iw
dL/dt = I*dw/dt
Torque (T) = I*ang.acc. (a)
so
a = T/I = mglsin[90] / Ml²/12
= 12g/l
this is wrong so i used l=l/2 as this is the distance to the cm.
so a = 24g/l also wrong

what am i doing wrong?

 

[Jebus_Chris]

Calculate the moment of inertia about the pivot point using the parallel axis theorem.